Math, asked by ramya8642, 11 months ago

A driver leaves is city at 6 a.m. and another driver the city from the same place at 11:30 a.m. in the same direction . if the average speed of the first driver was 55 km/hour and the second driver 65 km/hour. find the time at which they will be together.​


LAKSHMINEW: @Ramya
ramya8642: A driver leaves is city at 6 a.m. and another driver the city from the same place at 11:30 a.m. in the same direction . if the average speed of the first driver was 55 km/hour and the second driver 65 km/hour. find the time at which they will be together
LAKSHMINEW: ^.^ see @gurvit answered
ramya8642: i couldn't understand about relative speed
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ramya8642: plz explain me in detail
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Answers

Answered by Anonymous
35

Answer:

A driver leaves is city at 6 a.m

before another driver starts from the city first driver travels x kms

s=d/t

55=d/5.5hrs

d=302.5 kms

meeting time of both the drivers= distance/relative speed

time=302.5/7

43.21hrs

answer=11.30 am + 43.21 hours


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Answered by artiakshat888790
4

Step-by-step explanation:

Explanation

Lets assume, they will be together in 't' hours after 6 a.m. when the first driver leaves the city.

As the second driver leaves at 11:30 a.m., that means he leaves (11:30am - 6am)= 5.5 hours after the first driver. So, the time duration for second driver is (t- 5.5) hours.

The average speed of the first driver was 55 km/hour and the second driver 65 km/hour.

As Distance= Speed × Time , so the equation will be ....

\begin{gathered}55t=65(t-5.5)\\ \\ 55t=65t-357.5\\ \\ 55t-65t=-357.5\\ \\ -10t=-357.5\\ \\ t=\frac{-357.5}{-10}=35.75\end{gathered}

55t=65(t−5.5)

55t=65t−357.5

55t−65t=−357.5

−10t=−357.5

t=

−10

−357.5

=35.75

So, they will be together after 35.75 hours or 35 hours 45 minutes from 6 am , which is at (6 am+ 35 hours 45 minutes)= 5:45 p.m. on the next day.

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